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3 years ago

13

Students in Mrs. Walsh’s math class wanted to know the favorite snacks of students in the school. How could they generate a rand

om sample to survey?

1 answer:

frozen [14]3 years ago

6 0

Answer:

Step-by-step explanation:

Abajkavsowvsj some people who have never in the game and have a lot

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Cumulative GPA (CGPA) of a student is defined as the sum of the product of all grades multiplied by the credits of the correspon

user100 [1]

Answer:

probably about 3.5

Step-by-step explanation:

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2 years ago

Given the figure. Find AC.

Angelina_Jolie [31]

Image is missing, so i have attached it.

Answer:

AC = 10sin 40°

Step-by-step explanation:

From the diagram attached, using terms in trigonometric ratio, AC is the opposite side, BC is the adjacent side and AB is the hypotenuse.

Thus, since we want to find AC;

We know that in trigonometric ratios; opposite/hypotenuse = sin θ

In the diagram, θ = 40° and AB = 10

Thus,

AC/10 = sin 40°

Multiply both sides by 10 to get;

AC = 10sin 40°

7 0

4 years ago

What is the number when 0.83 is written as a fraction with a denominator of 100?

Grace [21]

The fraction would be 83/100

7 0

3 years ago

Read 2 more answers

A mountain climber needs to descend 3,435 feet. He wants to do it in 4 equal descents. How far should he travel in each descent?

kaheart [24]

Answer:

858.75

3,435/4=858.75

6 0

3 years ago

Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check al

Alla [95]

Let's rewrite each equation in the Slope-Intercept Form of the Equation of a Line. First, let's start with the main equation:

$\bullet \ 5x-2y=-6 \therefore y=\frac{5}{2}x+3$

Then, our options are the following:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}$

For two perpendicular lines it is true that the product of its slopes is:

$m_{1}m_{2}=-1$

$m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}$

According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

$y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!$

Finally, the right answer are:

$A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \\ \\ D) \ y+4=-\frac{2}{5}(x-5)$

8 0

3 years ago

Read 2 more answers